Unit 7 Stoichiometry. Chapter 12

Transcription

1 Unit 7 Stoichiometry Chapter 12

2 Objectives 7.1 use stoichiometry to determine the amount of substance in a reaction 7.2 determine the limiting reactant of a reaction 7.3 determine the percent yield of a reaction

3 7.1 Stoichiometry Definition The quantitative relationship that exists between the reactants and products of a chemical reaction and provides the basis for determining one amount from a provided amount.

4 Stoichiometry Definition Stoichiometry requires an understanding of chemical reactions (Unit 7) and mole conversions (Unit 8). To practice some of those skills, scroll to the end of this tutorial: Unit 7 Practice Unit 8 Practice

5 Solving a stoichiometry problem Every stoichiometry problem can be solved using the following four steps: 1. Balance the chemical equation, 2. Convert given amount to moles, 3. Use the mole ratio, 4. Convert your answer to the desired units. Steps 1, 2, and 4 are skills learned from Unit 7 and 8. Only step 3 is new.

6 The Mole Ratio Look at the balanced chemical reaction below. Compare the masses of each part. Convert those masses to moles for each chemical. Notice how the moles match the coefficients on the equation while the masses do not. 1 Ca + 1 Cl 2 1 CaCl g 70.9 g g 1 mol 1 mol 1 mol

7 The Mole Ratio The mole ratio provides the mathematical relationship between two parts of a chemical reaction. This relationship is given in the balanced chemical reaction. The masses do not match this equation. Anytime a change is required from one chemical to the next, both components must be in units of MOLES.

8 Solving Stoichiometry Problems Earlier, it was stated that four steps can solve stoichiometry problems. Of those steps, two of them can be skipped if the correct conditions apply. Steps 1 and 3 WILL always be used. Steps 2 and 4 can be skipped if: Skip step 2 if you are given moles at the beginning. Skip step 4 if the problem asks for moles. Practice

9 Steps to Stoichiometry Practice Which steps would you use? A + B C + D For the reaction above, you have 2 moles of A, how many moles of C will you get? For the reaction above, you have 2 grams of A, how many moles of C will you get? For the reaction above, you have 2 moles of A, how many grams of C will you get? For the reaction above, you have 2 grams of A, how many grams of C will you get?

10 Steps to Stoichiometry Practice Return Which steps would you use? A + B C + D For the reaction above, you have 2 moles of A, how many moles of C will you get? 1,3 For the reaction above, you have 2 grams of A, how many moles of C will you get? 1,2,3 For the reaction above, you have 2 moles of A, how many grams of C will you get? 1,3,4 For the reaction above, you have 2 grams of A, how many grams of C will you get? 1,2,3,4

11 Solving Stoichiometry Problems The steps to stoichiometry can be solved in a few ways. This tutorial will provide two methods for solving stoichiometry problems. The dimensional analysis (bridge) method The box method

12 Bridge Method The bridge method is the most common method used to solving stoichiometry problems and is set up below: Step 1: Balance the chemical equation Formula Mass given amount x 1 mole of given Formula Mass grams of given Step 2 : Convert to moles Step 3: Use the mole ratio Step 4: Convert to desired units x coefficient of wanted coefficient of given x grams of wanted 1 mole of wanted

13 The Box Method The box method is organized off of the balanced equation and always begins with the given amount. The setup appears below: Step 1 A + B C + D grams of given grams of wanted Step 2 formula mass of given formula mass of wanted moles of given mole ratio moles of wanted Step 4 Step 3

14 Solving a Stoichiometry Problem Oxgyen reacts with methane in a combustion reaction. If there was 5.0 moles of oxygen, how many moles of water are produced? Step 1: Balance the equation. CH 4 + 2O 2 CO 2 + 2H 2 O At this point, choose either the bridge or box method.

15 Bridge Method CH 4 + 2O 2 CO 2 + 2H 2 O given amount 1 mole of given coefficient of wanted grams of wanted x x x grams of given coefficient of given 1 mole of wanted 5 moles of O 2 x 2 moles H 2 O 2 moles O 2 = 5 moles of H 2 O Steps 2 and 4 are skipped because the problem gave moles and asked for moles. Return

16 Box Method CH 4 + 2O 2 CO 2 + 2H 2 O Step 2 Skipped 5.0 moles 2/2 5.0 moles Step 4 Skipped Return Steps 2 and 4 were skipped because the problem gave moles and asked for moles.

17 The answer to the problem was 5.0 moles of water. That problem is considered to be a mole to mole problem. The next problem is a gram to gram problem and will require all four steps.

18 Solving a stoichiometry problem When lead (ii) nitrate reacts with potassium iodide, potassium nitrate and lead (ii) iodide are produced. If you began with 120 grams of KI, how much PbI 2 would be produced? Step 1: Balance the equation Pb(NO 3 ) 2 + 2KI 2KNO 3 + PbI 2 Box or Bridge

19 Bridge Method Pb(NO 3 ) 2 + 2KI 2KNO 3 + PbI 2 given amount 1 mole of given coefficient of wanted grams of wanted x x x grams of given coefficient of given 1 mole of wanted 120 grams KI x 1 mole KI grams x 1 mole PbI 2 2 mole KI Formula Mass x 461 grams 1 mole PbI 2 = 170 grams PbI 2 Formula mass does not include the coeffecient Return

20 Box Method Pb(NO 3 ) 2 + 2KI 2KNO 3 + PbI grams 170 grams grams 461 grams 0.72 moles 1/ moles Return

21 The answer to the previous slide was 170 grams of lead (ii) iodide. Those are the basic mass stoichiometry problems. There will be problems that begin with moles and ask for grams and vice versa but they simply requiring dropping the appropriate step.

22 7.2 Limiting Reactants A chemical reaction is rarely setup to have the perfect amounts of each reactant. Generally, one will run out first. This reactant is called the limiting reactant. Generally, one is left over. This reactant is called the excess reactant. To get an idea of how this works, take a look at the next slide.

23 Limiting reactants Imagine the following reaction and assume that there is 4 moles of sodium and 4 moles of chlorine: 2Na + Cl 2 2NaCl

24 Limiting reactants As the reaction progresses, each mole of chlorine will use up two moles of sodium. 2Na + Cl 2 2NaCl

25 Limiting reactants Eventually, the sodium will run out. 2Na + Cl 2 2NaCl

26 Limiting reactants When that happens, there are still 2 moles of chlorine left over. This is our excess reactant 2Na + Cl 2 2NaCl

27 Calculating the limiting reactant In a problem like the one represented on the previous slides, it is possible to calculate the limiting reactant. One way to do this is to take one of the reactants and calculate how much of the other reactant would be required.

28 Calculating the limiting reactant In the problem from before, there was 4 moles of sodium and 4 moles of chlorine. If we take the 4 moles of sodium and calculate the amount of chlorine required, we would get the following: 2Na + Cl 2 2NaCl 4 moles x 1 mol Cl 2 2 mol Na = 2 moles Cl 2 Since we have 4 moles of Cl 2, we have enough to completely use up the sodium. Therefore, we know sodium will run out first and is thus the limiting reactant.

29 Calculating the limiting reactant Since we know that sodium is the limiting reactant, that means the chlorine is the excess reactant. Once we have determined the limiting reactant, we use its amount to calculate any other amount we would like to know. Amount of NaCl made: 4 mol Na x 2 mol NaCl 2 mol Na = 4 mol NaCl Amount of Cl 2 used: 4 mol Na x 1 mol Cl 2 2 mol Na = 2 mol Cl 2

30 7.3 Percent Yield Not all chemical reactions will undergo their reaction perfectly. When a reaction is performed, there are several places for error to occur. The percent yield is a evaluation of how efficient the reaction progresses as well as how well it was carried out by the scientist. A perfect percent yield would be 100% but is rarely achieved.

31 Percent Yield To calculate the percent yield, two values are required. Actual yield: amount that was achieved in the experiment Theoretical yield: calculated amount that should be achieved The mathematical calculation is as follows: actual yield theoretical yield x 100 = percent yield

32 Percent yield The actual yield is typically provided in the problem or determined in the lab. The theoretical yield will have to be calculated for each problem. Typically, this will be a gram to gram stoichiometry problem.

33 Percent yield-typical problem Calcium chloride reacts with oxygen to create calcium chlorate. If there was 20. grams of calcium chloride, what would the percent yield be if 30. grams calcium chlorate were produced in lab? When looking at this problem, we know the actual yield is 30. grams. To determine the theoretical, we will have to perform a stoichiometry problem to determine how much calcium chlorate we should have produced.

34 Percent yield Since we need to know how much Ca(ClO 3 ) 2 is supposed to be produced, a stoichiometry problem is required. 2CaCl 2 + 3O 2 2Ca(ClO 3 ) 2 20 g CaCl 2 x 1 mol CaCl g x 2 mol Ca(ClO 3) g x = 37 g Ca(ClO 2 mol CaCl 2 1 mole Ca(ClO 3 ) 3 ) 2 2 Box Method

35 Box Method-Percent yield 2CaCl 2 + 3O 2 2Ca(ClO 3 ) grams 37 grams grams grams 0.18 mol 2/ mol Return

36 Percent Yield Since we know the actual yield and theoretical yield, we are ready to calculate the percent yield. actual yield theoretical yield x 100 = percent yield 30. grams 37 grams x 100 = 81%

37 Percent yield A percent yield of 81% is actually a decent yield. Some chemical reactions do not work any better than an 81% yield. When working in the lab, it will be important to carefully watch the procedure and make notes on where you may lose product. This will allow you to explain why a 100% yield was not achieved.

38 Percent yield One problem that can appear in lab is a percent yield greater than 100%. This means that the final mass was more than theoretically possible. This will occur if there is a contaminant in with the product. There are two options at this point: Run the product through the isolation process again. Redo the experiment and report that the trial with over 100% was not used because it is not accurate.

39 This concludes the tutorial on measurements. To try some practice problems, click here. To return to the objective page, click here. To exit the tutorial, hit escape.

40 Unit 5 Practice Predict the products and balance the following equations: Na + Cl 2 CaCl 2 + BaSO 4 C 3 H 8 + O 2 CaClO 3 K + AuF 3

41 Unit 5 Practice: Answers Return Predict the products and balance the following equations: _2_Na + _1_Cl 2 2NaCl Synthesis _1_CaCl 2 + _1_BaSO 4 BaCl 2 + CaSO 4 Double replacement _1_C 3 H 8 + _5_O 2 3CO 2 + 4H 2 O Combustion _2_Ca(ClO 3 ) 2 3CO 2 + 2CaCl 2 Decomposition _1_K + _1_AuF 3 3KF + Au Single Replacement

42 Unit 6 Practice: Perform the following conversions: 34 grams of Cl 2 to moles 5.2 moles of Fe to grams 1.25 moles of CO 2 to grams 7390 grams of CaSO 4 to moles

43 Unit 6 Practice: Answers Return Perform the following conversions: 34 grams of Cl 2 to moles 34 grams x 1 mole 70.9 grams = 0.48 moles 5.2 moles of Fe to grams 5.2 moles x grams 1 mole = 290 grams 1.25 moles of CO 2 to grams 1.25 moles x grams 1 mole = 55.0 grams 7390 grams of CaSO 4 to moles 7390 grams x 1 mole grams = 54.3 moles

44 Definitions-Select the word to return to the tutorial Formula Mass: the mass of a chemical; calculated by adding the individual masses of each element within the chemical